Wastewater

Wastewater treatment protects rivers and health. Learn sewage characteristics (BOD, COD, SS, nutrients), BOD kinetics, primary, secondary (activated sludge, trickling filters) and tertiary treatment, F/M ratio, HRT, SRT, sludge handling and reuse.

Every litre of water a city uses comes back as wastewater carrying organic matter, nutrients and pathogens. Discharged untreated, it depletes oxygen in rivers and spreads disease.

Treatment plants harness microorganisms to eat the organic pollution, then separate the resulting biomass (sludge) from clean effluent.

An analogy

An activated sludge tank is a microbial restaurant: sewage is the food, air is the oxygen supply, and engineers balance food and diners (F/M ratio) so everything gets eaten.

What's in sewage

Domestic sewage: BOD 200–300 mg/L, suspended solids 200–400 mg/L, nitrogen, phosphorus, pathogens, plus industrial pollutants. BOD measures oxygen microbes need to decompose organics; COD measures chemically oxidisable matter (always ≥ BOD).

Treatment stages

Preliminary (screens, grit chambers), primary (sedimentation removes ~30 % BOD, ~60 % SS), secondary biological treatment (activated sludge, SBR, MBBR, trickling filters, UASB — removes 85–95 % BOD), tertiary (nutrient removal, filtration, disinfection) for reuse or sensitive waters.

Activated sludge design

Aeration tank + clarifier with sludge recycle. Key parameters: F/M 0.1–0.4 d⁻¹, MLSS 2000–4000 mg/L, HRT 4–8 h, sludge age (SRT) 5–15 days. Aeration uses ~50–60 % of plant energy.

Sludge, nutrients and resource recovery

Sludge is thickened, digested anaerobically (producing biogas for energy), dewatered and reused or disposed. Biological nitrogen removal (nitrification–denitrification) and phosphorus removal protect waters from eutrophication. Modern plants are 'water resource recovery facilities' producing reclaimed water, energy and fertiliser.

Formulas

• BOD kinetics: BOD_t = L₀ (1 − e^(−kt))
• Organic load: kg BOD/d = Q (MLD) × BOD (mg/L)
• F/M ratio: F/M = Q S₀ / (V X)
• HRT: θ = V / Q
• Removal efficiency: η = (S₀ − S)/S₀ × 100 %

Worked example: BOD₅ from ultimate BOD

Given: L₀ = 300 mg/L, k = 0.23 d⁻¹

Find: BOD₅.

• BOD₅ = 300 × (1 − e^(−1.15)) = 205 mg/L

Answer: ≈ 205 mg/L

Worked example: Activated sludge check

Given: Q = 10 MLD, S₀ = 200 mg/L; Aeration volume 4000 m³, MLSS 3000 mg/L

Find: Organic load, F/M, HRT.

• Load = 10 × 200 = 2000 kg BOD/d
• F/M = 10 000 × 200/(4000 × 3000) = 0.167 d⁻¹
• HRT = 4000/10 000 d = 9.6 h

Answer: 2000 kg/d, F/M 0.17, HRT 9.6 h

Common mistakes

• Wrong: Designing on average flow without peak hydraulic checks. — Right: Clarifiers and channels must pass peak wet-weather flows.
• Wrong: Ignoring sludge handling. — Right: Sludge treatment is 30–50 % of plant cost — design it upfront.

Applications

• Municipal STPs: Activated sludge, SBR and MBBR plants for cities and towns.
• Industrial effluent: Anaerobic (UASB) pretreatment of high-strength wastes.
• Water reuse: Tertiary treatment for irrigation, industry and indirect potable reuse.

Key takeaways

• BOD_t = L₀(1 − e^(−kt)).
• Load = Q × BOD; F/M = QS₀/(VX).
• Primary → secondary (biological) → tertiary.
• Handle sludge and recover resources.

Going further

• Streeter–Phelps: Models the oxygen sag in rivers downstream of discharges.
• Membrane bioreactors: MBRs replace clarifiers with membranes for compact, high-quality effluent.
• Energy-neutral plants: Biogas CHP, efficient aeration and thermal hydrolysis make plants self-powered.